For a stellite orbiting close to the surface of earth the period of revolution is `84 min`. The time period of another satellite orbiting at a geight three times the radius of earth from its surface will be

Options

A. `(84) 2 sqrt(2) min`
B. `8 (84) min`
C. `(84) 3sqrt(3) min`
D. `3 (84) min`

Answer

B

Explanation

`T prop r^(3//2)`
`(T_(2))/(T_(1)) = ((r_(2))/(r_(1)))^(3//2) = ((3R + R)/(R))^(3//2) = 8`
`:. T_(2) = 8T_(1)`